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Why must the upper limit match when integrating?

  1. Nov 23, 2013 #1
    For example, if you have the function f(x) = x2 then find:

    d/dx any number3x∫ t2dt

    Why must the dx in d/dx ∫f(t)dt always match the upper limit in order to compute the integral? Why is the lower limit of no concern? I know that you must take chain rule into consideration and change 3x to u, and then do du/dx but why does the upper limit only matter?

    Any help would be great!
  2. jcsd
  3. Nov 23, 2013 #2


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    Huh I don't understand your question?
  4. Nov 23, 2013 #3
    When applying the fundamental theorem of calculus, think about everything in terms of differentiation.

    Say [itex]f:\mathbb R\to\mathbb R[/itex] is some continuous, bounded function. For any [itex]a,b\in \mathbb R[/itex], let [itex]F(a,b)=\int_a^bf(x)\text{d}x[/itex].

    The FTC tells us that differentiating [itex]F(a,b)[/itex] by [itex]b[/itex] gives [itex]f(b)[/itex] (no matter what [itex]a[/itex] is), and differentiating [itex]F(a,b)[/itex] by [itex]a[/itex] gives [itex]-f(a)[/itex] (no matter what [itex]b[/itex] is).

    As an example, think about the case where [itex]f(x)>0[/itex] is your speed at time [itex]x[/itex], in which case [itex]F(a,b)[/itex] (for [itex]a<b[/itex]) is just how far you've traveled between time [itex]a[/itex] and time [itex]b[/itex].

    Talking about how fast you're going at time [itex]b[/itex] doesn't require any information about when you started moving (i.e. time [itex]a[/itex]).
  5. Nov 23, 2013 #4


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    You are making some assumptions that are unjustified- in particular that it is the upper
    limit that is important!

    In order to be able to differentiate with respect to x, you must have a function of x. One way to do that is to have one or the other limit of integration a function of x.
    Both [itex]\int_0^x t^2 dt= (1/3)x^3[/itex] and [itex]\int_x^1 e^t dt= e- e^x[/itex] are differentiable with respect to x.

    Yet another is to have constant limits of integration while the integrand is a function of both x and the "variable of integration:
    [itex]\int_0^1 e^{x+ t}dt= e^x\int_0^1 e^t dt= e^x[e- 1][/itex] is differentiable with respect to x.

    (Strictly speaking, of course, a the integral does NOT have to have an "x" anywhere in order to be differentiable with respect to x! Of course, then, the derivative is 0.)
    Last edited by a moderator: Nov 24, 2013
  6. Nov 23, 2013 #5


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    To elaborate on what Halls said what is important is the places where the variable appears. The general rule is called Leibniz integral rule and states that
    $$\dfrac{d}{dt} \int_{\mathrm{a}(t)}^{\mathrm{b}(t)} \mathrm{f}(x,t) \, \mathrm{d}x = \int_{\mathrm{a}(t)}^{\mathrm{b}(t)} \mathrm{f} ^{(0,1)}(x,t) \, \mathrm{d}x + \mathrm{f}(\mathrm{b}(t),t)\mathrm{b}^\prime (t)-\mathrm{f}(\mathrm{a}(t),t)\mathrm{a}^\prime (t) \\ \text{where } \mathrm{f} ^{(0,1)}(x,t) \text{ is the derivative of f with respect to t with x treated as constant.}$$
  7. Nov 23, 2013 #6
    ^ This.

    And Leibniz's rule is really just what you get by combining FTC and the chain rule.
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